Question

A boat is traveling along a circular curve having a radius of 100 ft. If its speed at t = 0 is 15 ft>s and is increasing at v # = (0.8t) ft>s 2 , determine the magnitude of its acceleration at the instant t = 5 s.

Answer 1

Answer:

a=7.42 ft/s²

Explanation:

Given data

Radius r=100 ft

Speed v=15 ft/s at t=0

acceleration a₁=(0.8t) ft/s²

To find

acceleration a

solution

As acceleration is first derivative of velocity with respect to time

So

$a=\frac{dV}{dt}\\ dV=a.dt\\integrate\\\int\limits{} \, dv=\int\limits{a} \, dt\\v-v_{o} =\int\limits{0.8t} \, dt\\v-15=0.4t^{2}\\ v=0.4t^{2}+15\\ at\\t=5s\\v=0.4(5)^{2}+15\\v=25ft/s$

As

$velocity=distance/time\\time=distance/velocity\\time=100ft/25ft/s\\time=4s$

a₂=v/t

a₂=(25ft/s)/4s

a₂=6.25 ft/s²

$a=\sqrt{(6.25)^{2}+(0.8*5)^{2} }\\ a=7.42 ft/s^{2}$

Answer 2

The acceleration of the Boat at time t=5 sec will be $a=7.42 \dfrac{ft}{s^2}$

What will be the acceleration of the boat?

It is given that

Radius r=100 ft

Speed      $V=15\dfrac{ft}{sec}$

acceleration $a_{1}=(0.8t)\dfrac{ft}{sec}$  

Now to find the acceleration at an instant t=5 sec

We know that

$a=\dfrac{dV}{dt} \ \ or \ \ dV=a.dt$

$\int\limits {} \, dV=\int\limits {a} \, dt$

$V-V_{0}=\int\limit{0.8t} \, dt$

$V-15=0.5t^2$

$V=0.4t^{2}+15$

At t=5sec

$V=0.4(5^2)+15$

$V=25\dfrac{ft}{sec}$

Now

$a_{2}=\dfrac{v}{t} =\dfrac{25}{4} =6.25\dfrac{ft}{sec^2}$

$a=\sqrt{a_{1}^{2} +a_2^{2}$

$a=\sqrt{6.25^{2}+(0.8\times5)^2}$

$a=7.42\dfrac{ft }{sec^2}$

Thus the acceleration of the Boat at time t=5 sec will be $a=7.42 \dfrac{ft}{s^2}$

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